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Suppose that $f(x) = o(\sqrt{x})$ as $x\rightarrow\infty$ and let $x^*(a)$ denote the minimizer of $f(x) + a^{3/2}/x$, that is, the value of $x$ that minimizes said expression (assuming such a value exists). As $a\rightarrow\infty$, is it true that $x^*(a) = \omega(a)$, i.e. that the minimizer grows super-linearly in $a$?

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What is a "minimizer" in this context? – Henning Makholm Jul 15 '12 at 1:29
The value of $x$ that minimizes $f(x) + a^{3/2}/x$. Of course, the expression may not have a minimizer (e.g. if f(x) = 0) – Charlie Jul 15 '12 at 1:30
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Yes. It suffices to show that the minimum is $o(\sqrt{a})$. Suppose not. Then there is $c>0$ and a sequence $a_n\to\infty$ such that $f(x)+a_n^{3/2}/x\ge c\sqrt{a_n}$ for all $n$ and all $x$. But for large $x$ we have $f(x)<c^{3/2}\sqrt{x}/1000$. Setting $x=100a_n/c$ yields a contradiction.

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